if f(x) = x-3/(x+4)(x^2-2), express f(x) in partial fraction
x-3/(x+4)(x^2-2) = A/(x+4) + (Bx+C)/(x^2-2)
x-3 = A(x^2-2) + (Bx + C)(x+4)
if x = -4
-4-3 = A((-4)^2-2) + (B(-4) + C)(-4 + 4)
-7 = 14A, A=-1/2
if x=0,
0-3 = A(-2) + (B(0) + C)(4)
-3 = -2A + 4C
-3 = -2(-1/2) + 4C
-3 +1 = 4C
C= -1/2
if x=2
2-3 = A(2) + (B(2) + C)(6)
-1 = (-1/2)(2) + 12B + 6(-1/2)
-1 = -1 +12B -3
-1+4 = 12B
B = 1/3
therefore, x-3/(x+4)(x^2-2) = -1/2(x+4) + ((1/4)(x) + (-1/2))/(x^2-2)
= -1/2(x+4) + (x-2)/4(x^2-2)
This is the final answer i get, but some how in the text book the answer given is
= -1/2(x+4) + (x-2)/2(x^2-2)
Do you mind to let me know is my did wrongly or the answer provided is wrong.
thank you.